Rewrite the function by completing the square. $f(x)=x^{2}-16x+76$ $f(x)=(x+$
Solution: We want to complete $x^2{-16}x$ into a perfect square. To do that, we should add $\left(\dfrac{{-16}}{2}\right)^2={64}$ to it: $x^2{-16}x+{64}=(x-8)^2$ In order to keep the expression equivalent, we add and subtract ${64}$, not forgetting the expression's constant term, $76$ : $\begin{aligned} f(x)&=x^2-16x+76 \\\\ &=x^2-16x+{64}+76-{64} \\\\ &=(x-8)^2+76-64 \\\\ &=(x-8)^2+12 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x - 8)^2 +12$ This is equivalent to $f(x)=(x+{-8})^2+12$